If we're given V=2x^2+2y^2 and we're told to find the field at a point x=3, y=3, would this method be correct?
Ex=d(2x^2)/dx=4x=12
Ey=d(-2y^2)/dy=-4y=-12
Therefore, E={12i - 12j}
Not entirely sure, this is the first of its kind.
If you are referring to the $\displaystyle \nabla$ in the topsquark's expression $\displaystyle E = - \nabla V,$ then it's not a Greek letter, actually. It's called the "del" operator, or also the "nabla". In this context, it indicates that you are to take the gradient of the scalar function $\displaystyle V.$
Not true for your problem, actually. If you were working in only one dimension, it would be true. For your case, you should have the vector equationAlright, I have yet to encounter that sign so it's probably best if I say E=-dv/dx correct?
$\displaystyle \mathbf{E}=-\left\langle\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y},\dfrac{\partial V}{\partial z}\right\rangle.$
Or, for two dimensions, you'd have
$\displaystyle \mathbf{E}=-\left\langle\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y}\right\rangle.$